Suspension of a product - tricky homotopy equivalence

Let $(X,x_0), (Y,y_0)$ be well-pointed spaces (inclusion of the basepoints is a cofibration). Show the following homotopy equivalence $$ \Sigma (X\times Y) \simeq \Sigma X \lor \Sigma Y \lor \Sigma (X\land Y), $$ where $\Sigma$ means a suspension (or reduced suspension if one prefers since it doesn't matter for well-pointed spaces) and $\land$ is a smash product.

Assuming we are using reduced suspension, it is quite clear that $\Sigma X \lor \Sigma Y$ is a subspace of the lhs, but I don't know how to somehow pull it outside and get $\Sigma (X\land Y)$.

Edit: I know the homotopy equivalence can be deduced from general theorems on "homotopy functors" (I hope that's how they are called in English) from chapter 7.7 in Spanier. But I was told there is an explicit proof and that's the one I'm looking for.

I don't know how explicit we can go, but I'll give it a try. We have to go first through the homotopy-theoretical part.

Since $\{ * \} \subseteq X, \{ * \} \subseteq Y$ are cofibrations, $X \vee Y \subseteq X \times Y$ also is. Let $Z$ be a pointed space and consider the long exact sequence of homotopy for the pair $X \vee Y \subseteq X \times Y$, ie. the sequence

$\ldots \rightarrow [\Sigma ^{2}(X \vee Y), Z] \rightarrow [\Sigma(X \wedge Y), Z] \rightarrow [\Sigma(X \times Y), Z] \rightarrow [\Sigma(X \vee Y), Z] \rightarrow [X \wedge Y, Z] \rightarrow \ldots$,

where $[-,-]$ is the pointed set of homotopy classes of basepoint-preserving maps. Note that for any $n \geq 0$, $\Sigma^{n}(X \vee Y)$ is homeomorphic to $\Sigma^{n}X \vee \Sigma ^{n} Y$. I will not distinguish between the two.

Let $k \geq 1$ and define a map

$\psi ^{k}: \Sigma^{k}(X \times Y) \rightarrow \Sigma^{k}X \vee \Sigma^{k}Y$

$\psi ^{k} = \Sigma^{k}(i_{X} \pi_{X}) + \Sigma^{k}(i_{Y} \pi_{Y})$,

where $\pi: X \times Y \rightarrow X, Y$ are the projections and $i: X, Y \rightarrow X \vee Y$ are the inclusions. Addition is performed via the suspension structure on $\Sigma^{k}(X \times Y)$, so this is why we require $k \geq 1$. (Observe that even though I denote it by addition this is not necessarily commutative for $k=1$.)

If $j: X \vee Y \hookrightarrow X \times Y$ is the inclusion, then I claim that $\psi ^{k}$ is the left inverse to $\Sigma^{k}j$, ie. $\psi ^{k} \circ \Sigma^{k}j = id_{\Sigma^{k}(X \vee Y)}$. This is important because $\Sigma^{k}j$ are connecting maps in the long exact sequence of homotopy. Indeed, one computes

$\psi ^{k} \circ (\Sigma^{k}j) = (\Sigma^{k}(i_{X} \pi_{X}) + \Sigma^{k}(i_{Y} \pi_{Y})) \circ \Sigma^{k}j = \Sigma^{k}(i_{X} \pi_{X} j) + \Sigma^{k}(i_{Y} \pi _{Y} j) = \Sigma^{k}(id_{X} \vee const) + \Sigma^{k}(const \vee id_{Y}) \simeq (\Sigma^{k}id_{X} + const) \vee (const + \Sigma^{k}id_{Y}) \simeq \Sigma^{k}id_{X} \vee \Sigma^{k}id_{Y} \simeq id_{\Sigma^{k}X \vee \Sigma^{k}Y}$.

(One can also see this geometrically.) This immediately implies that for all $k \geq 1$ and all $Z$ the $[\Sigma^{k}(X \times Y), Z] \rightarrow [\Sigma^{k}(X \vee Y), Z]$ induced by $j$ is surjective and - by exactness of the long exact sequence - that for all $n \geq 1$ the map $[\Sigma^{n}(X \smash Y), Z] \rightarrow [\Sigma^{n}(X \times Y), Z]$ has zero kernel. In particular, for $k=1$ we have the short exact sequence of groups

$0 \rightarrow [\Sigma(X \wedge Y), Z] \rightarrow [\Sigma(X \times Y), Z] \rightarrow [\Sigma(X) \vee \Sigma(Y), Z] \rightarrow 0$

Moreover, the map induced by $\psi^{1}$ splits it and shows that there is a natural isomorphism

$\phi: [\Sigma(X \wedge Y), Z] \rtimes [\Sigma(X) \vee \Sigma(Y), Z] \rightarrow [\Sigma(X \times Y), Z]$,

of groups, where the product is only semi-direct, because our groups are not necessarily abelian. This is enough for our purposes, since we also have natural bijections

$[\Sigma(X \wedge Y), Z] \rtimes [\Sigma(X) \vee \Sigma(Y), Z] \simeq [\Sigma(X \wedge Y), Z] \times [\Sigma(X) \vee \Sigma(Y), Z] \simeq [\Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y), Z]$.

(The second one follows from from the fact that $\vee$ is the direct sum in the category of pointed spaces.) Yoneda lemma establishes that there is an isomorphism

$\theta: \Sigma(X \times Y) \rightarrow _{\simeq} \Sigma(X \smash Y) \vee \Sigma(X) \vee \Sigma(Y) $

in the homotopy category of pointed spaces, ie. a homotopy equivalence that we were after. It takes a little bookkeeping in the above Yoneda-lemma argumentation to see that such map is given by

$\theta = \Sigma(p) + \psi^{i} = \Sigma(p) + \Sigma^{1}(i_{X} \pi_{X}) + \Sigma^{1}(i_{Y} \pi_{Y})$,

where $p: X \times Y \rightarrow X \wedge Y$ is the natural projection. (This is what we get if we start with $id \in [\Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y), \Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y)]$ and trace it back by all the bijections above to $[\Sigma(X \times Y), \Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y)]$ - and this is the way to discover the isomorphisms "hidden" by Yoneda lemma.)

I understand that my exposition is far from perfect, but if you would like me to go into more detail over some parts, please comment.

This is proposition 4I.1 in Hatcher. The argument given there is geometric and elementary. It goes as follows:

We assume $X$ and $Y$ are CW complexes and we use the reduced suspension everywhere (which in this case is homotopy equivalent to the free suspension). I assume that your more general situation of well-pointed spaces works the same.

Consider the reduced Join $X*Y$ which is, like the regular join, the quotient of $X\times Y\times [0,1]$ by the relations

$$ (x,y_1,0)\sim(x,y_2,0) $$ $$ (x_1,y,1)\sim(x_2,y,1) $$

but we also collapse to a point the segment $(x_0,y_0,t)$. i.e. we collapse one "face" of the "cube" to $X$ and the opposite to $Y$ and the segment connecting the base points. Now, we glue in the cones on $X$ and $Y$ respectively along the collapsed faces and obtain a space

$$ Z=CX\sqcup_X (X*Y) \sqcup_Y CY $$

We will show that it is homotopy equivalent to the two spaces in your question.

On the one hand, if we collapse the two cones $CX$ and $CY$, what we get (by unwinding the definitions) is precisely the reduced suspension of $X\times Y$ (note that we used the reduced join). Since collapsing a contractible subcomplex does not change the homotopy type, we have a homotopy equivalence $Z\simeq \Sigma (X\times Y)$.

On the other hand, inside the reduced join $X*Y$, we have the spaces $X*y_0$ and $x_0*Y$ which are also contractible, as they are just cones on $X$ and $Y$ respectively, and collapsing them gives $\Sigma(X)\vee \Sigma(X\wedge Y)\vee \Sigma(Y)$. As before, this space is homotopy equivalent to $Z$ and we are done.